Question

In presence of $HC{l_{\left( {aq} \right)}}$,${H_2}S$ gas results the precipitation of group-2 cations but not of group-4 cations during qualitative analysis. It is due to.
A) Lower concentration of ${S^{2 - }}$
B) Higher concentration of ${S^{2 - }}$
C) Lower concentration of ${H^ + }$
D) Higher concentration of ${H^ + }$

Answer 1

Hint:We know that an effect that suppresses the ionization of an electrolyte when another electrolyte which contains ions in common which is present in the first electrolyte is called the common ion effect.
Complete step by step answer:
Hydrogen sulfide \[{H_2}S\] gas is passed in presence of \[HCl\] to analyze the cations of the second group. Due to the common ion effect, ionization of \[{H_2}S\] decreases and less Sulfide ions are obtained. These Sulfide ions are enough for the precipitation of second group cations in form of their Sulfides due to lower value of their solubility product \[\left( {{K_{{\text{s}}p}}} \right)\]
Therefore, the option A is correct.

Additional Note:
Solubility and common ion effect:
The common ion effect is used to get drinking water from aquifers containing chalk or limestone. In order to decrease the hardness of the water sodium carbonate is added.
The common ion effect is used to precipitate calcium carbonate from water through the addition of sodium carbonate.
The Calcium carbonate precipitated using sodium carbonate.
The soaps are precipitated by adding sodium chloride to the soap solution in order to reduce its solubility.

Note: Solubility Product:
Solubility product is defined as the equilibrium constant of a chemical reaction in which a solid ionic compound dissolves to yield its ions in solution.
The value of the solubility product usually increases with an increase in temperature due to increased solubility.
Some of the factors that affect the value of solubility products are the common-ion effect, the diverse-ion effect, and the presence of ion-pairs.
If the ionic concentration has more value than the solubility product, sufficient precipitate would be formed to decrease concentrations to give an answer similar to the solubility product.